RegEx: Grabbing values between quotation marks

Question

I have a value like this:

\"Foo Bar\" \"Another Value\" something else

What regex will return the

Answer 1

I liked Eugen Mihailescu's solution to match the content between quotes whilst allowing to escape quotes. However, I discovered some problems with escaping and came up with the following regex to fix them:

(['"])(?:(?!\1|\\).|\\.)*\1

It does the trick and is still pretty simple and easy to maintain.

Demo (with some more test-cases; feel free to use it and expand on it).

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_{PS: If you just want the content between quotes in the full match ($0), and are not afraid of the performance penalty use:}

(?<=(['"])\b)(?:(?!\1|\\).|\\.)*(?=\1)

_{Unfortunately, without the quotes as anchors, I had to add a boundary \b which does not play well with spaces and non-word boundary characters after the starting quote.}

_{Alternatively, modify the initial version by simply adding a group and extract the string form $2:}

(['"])((?:(?!\1|\\).|\\.)*)\1

_{PPS: If your focus is solely on efficiency, go with Casimir et Hippolyte's solution; it's a good one.}